# Homework Help: Infinite square well

1. Nov 4, 2016

### Vitani11

1. The problem statement, all variables and given/known data
A particle is in the n=1 state in an infinite square well of size L. What is the probability of finding the particle in the interval Δx = .006L at the point x = 3L/4?

2. Relevant equations
ψ(x) =√(2/L) sin(nπx/L)

3. The attempt at a solution
The problem states that because Δx is very small I don't need to do any integration. I've been integrating anyway because it's fun in this particular case- but my problem is what to do with x = 3L/4. I've integrated up to x= 0.006L in ψ2(x) and I'm getting the correct answer for if the particle was at x=L/2. What do I do here?

2. Nov 4, 2016

### BvU

You show your work and then we can comment . 'up to x = 0.006 L' makes me feel uneasy: up to x = 0.006 L from where ?

3. Nov 4, 2016

### Vitani11

So here is the integration from 0 to 0.006L. I tried the integration from 0.006L to 3L/4 also and im getting a number greater than 1...

File size:
26.2 KB
Views:
71
4. Nov 4, 2016

### Vitani11

Holy **** the answer was simply 0.006. can you please explain this to me?

5. Nov 4, 2016

### BvU

Disaster ! But the original exercise asks about probability to find the bugger in an interval of width 0.006 L at 3/L/4 , so I would be reassured if you integrated from 0.75 L to 0.7506 : or so. Since the sine varies only little in uch a small interval, you can just take the value. But, if I may ask: what do you integrate (what's the integrand) ? (edit: never mind, iI can see you do it right from the grey-on grey picture). Now about the value.

6. Nov 4, 2016

### Vitani11

I figured that since at L/2 there would be the highest probability at that point. At 3L/4 you're a little bit further right in the well and since the probability of finding a particle represents a sine wave then it would have to be a number smaller than what I got integrating from 0 to 0.006L (which was twice 0.006L) but how are you to know that it's at exactly 0.006L?

7. Nov 4, 2016

### BvU

Doesn't sound good to me, but I can be wrong....
 no, sounds good to me. Why ?

8. Nov 4, 2016

### Vitani11

The integrand is √(2/L) sin2 (nπx/L)

9. Nov 4, 2016

### BvU

You missed a factor 1/2: $\quad{\bf 2} \sin^2x = 1-\cos 2x$

 and the numerics look mistaken too: for small x , $\ x-1/2 \sin2x$ should be closer to 0.

10. Nov 4, 2016

### BvU

so at 3L/4 that sine squared is 1/2, and ${2\over L } {1\over 2 } 0.006 L = 0.006$

Simple exercise, lots to learn !. (not only you, me 2)

11. Nov 4, 2016

### Vitani11

I see. Thank you.